ABC is a triangle. The bisector of ∠C meets AB at D. Show that CD2 < CA·CB.
Solution
Extend CD to meet the circumcircle again at C'. Then ∠CAB = ∠CC'B, so triangles CAD and CC'B are similar. Hence CA/CD = CC'/CB. So CA·CB = CD·CC' > CD2.
© John Scholes
jscholes@kalva.demon.co.uk
1 Nov 2003
Last corrected/updated 1 Nov 03